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Questionsย andย Answersย Sheetย 2

ย 

ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย ย  ย Seriesย 

ย 

Questionย #1ย 

Aย pairย ofย diceย isย rolledย untilย aย sumย ofย eitherย 5ย orย 7ย appears.ย Findย theย probabilityย thatย aย 5ย occurs first.ย Hint:ย 
Letย 
ย 

๐ธ

๐‘›

ย ย 

denoteย theย eventย thatย aย 5ย occurs onย theย nthย rollย andย noย 5ย orย 7ย occurs onย theย firstย n-1ย rolls.ย Computeย 
ย 

๐‘ƒ(๐ธ

๐‘›

)

ย ย 

andย argueย thatย 
ย 

โˆ‘ย ๐‘›ย =ย 1

โˆž

๐‘ƒ(๐ธ

๐‘›

)

ย ย 

is theย desiredย probability.

ย 

Answer:ย 

Experiment:ย Throwingย twoย fairย sixย sidedย dice.ย 
Useย theย notationย fromย theย exercise:ย 
ย 

๐ธ

๐‘›

ย ย -ย Eventย thatย aย sumย ofย 5ย isย rolledย inย theย n-thย roll,ย andย neitherย 5ย orย 7ย wereย rolledย before.ย 

Nowย every eventย whereย theย sumย ofย 5ย is rolledย beforeย 7ย is someย ย 

๐ธ

๐‘›

ย ย -ย becauseย theย checkingย ifย this eventย 

happenedย inย theย experimentย is:ย fromย theย beggingย watchย forย sumย ofย 5ย orย 7,ย ifย theย sumย ofย 5ย appearedย firstย 
bothย sums,ย letsย sayย inย n-thย rollย thisย is eventย ย 

๐ธ

๐‘›

ย ย 

Weย haveย 36ย equallyย likely possibilities forย every diceย roll,ย this means aย totalย ofย ย 

36

๐‘›

ย ย possibilities.ย 

4ย ofย theseย ย 

36ย โˆ’ย (1,4),ย (4,1),ย (2,3),ย (3,2)

ย ย areย possibleย ifย theย sumย is 5ย andย 6ย areย possibleย if theย sumย is 7.ย 

Thisย leaves 26ย possibleย rolls thatย sumย upย toย aย numberย that's notย 5ย orย 7.ย 
If ย 

๐ธ

๐‘›

ย ย happened,ย theย firstย n-1ย rolls areย someย fromย ย 

26

๐‘›โˆ’1

ย ย vectors andย theย lastย oneย is oneย ofย 4ย rolls.ย 

This ciuntingย uses theย basic principleย ofย countingย thatย laterย translates toย indepence.ย 

ย 

๐‘ƒ(๐ธ

๐‘›

)ย =

26

๐‘›โˆ’1

โ‹…4

36

๐‘›

=

4

36

(

26

36

)

๐‘›โˆ’1

=

1

9

(

13

18

)

๐‘›โˆ’1

ย ย 

ย 

โˆ‘

๐‘ƒ(๐ธ

๐‘›

)

โˆž

๐‘›=1

=ย โˆ‘

1

9

(

13

18

)

๐‘›โˆ’1

โˆž

๐‘›=1

ย ย 

ย 

=

1

9

โˆ‘

(

13

18

)

๐‘˜

โˆž

๐‘˜=0

ย ย 

ย 

=

1

9

โ‹…

1

1โˆ’

13
18

ย ย 

ย 

=

1

9

โ‹…

1

5

18

ย ย 

ย 

=

1

9

โ‹…

18

5

ย ย 

ย 

=

2

5

ย ย 

Result:ย ย 

2

5

ย 

ย 

Questionย #2ย 

Proveย thatย ย 

1ย ยทย 1!ย +ย 2ย ยทย 2!ย +ยทยทยทย +๐‘›ย ยทย ๐‘›!ย =ย (๐‘›ย +ย 1)!ย โˆ’ย 1

ย wheneverย nย is aย positiveย integer.

ย 

Answer:ย 

Toย proofย ย 

1ย โ‹…ย 1!ย +ย 2ย โ‹…ย 2!ย +ย โ‹ฏย +ย ๐‘›ย โ‹…ย ๐‘›!ย =ย (๐‘›ย +ย 1)!ย โˆ’ย 1

ย ย forย every positiveย integerย n.ย 

Proofย byย inductionย 

Letย ย 

๐‘ƒ(๐‘›)

ย beย ย 

1ย โ‹…ย 1!ย +ย 2ย โ‹…ย 2!ย +ย โ‹ฏย +ย ๐‘›ย โ‹…ย ๐‘›!ย =ย 1ย โ‹…ย 1ย =ย (๐‘›ย +ย 1)!ย โˆ’ย 1

ย ย 

Basis stepย n=1ย 

ย 

1ย โ‹…ย 1!ย +ย 2ย โ‹…ย 2!ย +ย โ‹ฏย +ย ๐‘›ย โ‹…ย ๐‘›!ย =ย 1ย โ‹…ย 1!ย =ย 1ย โ‹…ย 1ย =ย 1

ย ย 

ย 

(๐‘›ย +ย 1)!ย โˆ’ย 1ย =ย (1ย +ย 1)!ย โˆ’ย 1ย =ย 2!ย โˆ’ย 1ย =ย 2ย โˆ’ย 1ย =ย 1

ย ย 

Weย thenย noteย P(1)ย is true.ย 

background image

Inductionย stepย letย ย 

๐‘ƒ(๐‘˜ย +ย 1)

ย ย isย alsoย trueย 

ย 

1ย โ‹…ย 1!ย +ย 2ย โ‹…ย 2ย +ย โ‹ฏย +ย ๐‘˜ย โ‹…ย ๐‘˜!ย =ย (๐‘˜ย +ย 1)!ย โˆ’ย 1

ย ย 

Weย needย toย proveย thatย ย 

๐‘ƒ(๐‘˜ย +ย 1)

ย ย isย alsoย true.ย 

ย 

1ย โ‹…ย 1!ย +ย 2ย โ‹…ย 2!ย +ย โ‹ฏย +ย ๐‘˜ย โ‹…ย ๐‘˜!ย +ย (๐‘˜ย +ย 1)ย โ‹…ย (๐‘˜ย +ย 1)!

ย ย 

ย 

=ย (๐‘˜ย +ย 1)!ย โˆ’ย 1ย +ย (๐‘˜ย +ย 1)ย โ‹…ย (๐‘˜ย +ย 1)!

ย ย 

ย 

=ย 1ย โ‹…ย (๐‘˜ย โ‹…ย 1)!ย โˆ’ย 1ย +ย (๐‘˜ย +ย 1)ย โ‹…ย (๐‘˜ย +ย 1)!

ย ย 

ย 

=ย 1ย โ‹…ย (๐‘˜ย +ย 1)!ย +ย (๐‘˜ย +ย 1)ย โ‹…ย (๐‘˜ย +ย 1)!ย โˆ’ย 1

ย ย 

ย 

=ย (1ย +ย ๐‘˜ย +ย 1)(๐‘˜ย +ย 1)!ย โˆ’ย 1

ย ย 

ย 

=ย (๐‘˜ย +ย 2)!ย โˆ’ย 1

ย ย 

ย 

=ย ((๐‘˜ย +ย 1)ย +ย 1)!ย โˆ’ย 1

ย ย 

Weย thenย noteย thatย ย 

๐‘ƒ(๐‘˜ย +ย 1)

ย ย is alsoย true

ย 

Questionย #3ย 

Findย theย Maclaurinย seriesย forย f(x)ย usingย theย definitionย ofย aย Maclaurinย series.ย [Assumeย thatย fย has aย powerย 

seriesย expansion.ย Doย notย showย thatย ย 

๐‘…๐‘›(๐‘ฅ)ย โ†’ย 0

ย .]ย Alsoย findย theย associatedย radius ofย convergence.ย ย 

๐‘“(๐‘ฅ)ย =

2

๐‘ฅ

ย 

Answer:ย 

Findย aย fewย derivatives,ย andย calculateย theirย values atย a=0.ย 

ย 

๐‘“(๐‘ฅ)ย =ย 2

๐‘ฅ

๐‘“(0)ย =ย 1

ย ย 

ย 

๐‘“(๐‘ฅ)ย =ย (lnย 2)2

๐‘ฅ

๐‘“

โ€ฒ

(0)ย =ย lnย 2

ย ย 

๐‘“(๐‘ฅ)ย =ย (lnย 2)

2

2

๐‘ฅ

๐‘“

โ€ฒโ€ฒ

(0)ย =ย (lnย 2)

2

ย ย 

ย 

๐‘“(๐‘ฅ)ย =ย (lnย 2)

3

2

๐‘ฅ

๐‘“

โ€ฒโ€ฒโ€ฒ

(0)ย =ย (lnย 2)

3

ย ย 

ย 

๐‘“

(4)

(๐‘ฅ)ย =ย (lnย 2)

4

2

๐‘ฅ

๐‘“

(4)

(0)ย =ย (lnย 2)

4

ย ย 

Plugย everythingย intoย theย Maclaurinย generalย formย 

ย 

๐‘“(๐‘ฅ)ย =ย ๐‘“(0)ย +

๐‘“

โ€ฒ

(0)

1!

๐‘ฅย +

๐‘“

โ€ฒโ€ฒ

(0)

2!

๐‘ฅ

2

+

๐‘“

โ€ฒโ€ฒโ€ฒ

(0)

3!

๐‘ฅ

3

+ย โ‹ฏ

ย ย 

ย 

๐‘“(๐‘ฅ)ย =ย 1ย +

lnย 2

1!

๐‘ฅย +

(lnย 2)

2

2!

๐‘ฅ

2

+

(lnย 2)

3

3!

๐‘ฅ

3

+

(lnย 2)

4

4!

๐‘ฅ

4

+ย โ‹ฏ

ย ย 

Findย theย patternย ofย theย numbers toย eriteย inย summationย formย 

ย 

๐‘“(๐‘ฅ)ย =ย โˆ‘

(lnย 2)

๐‘›

๐‘ฅ

๐‘›

๐‘›!

โˆž

๐‘›=0

ย ย 

Useย theย Ratioย Testย 

ย 

๐‘Ž

๐‘›

=

(lnย 2)

๐‘›

๐‘ฅ

๐‘›

๐‘›!

ย ย 

1

1

1

(lnย 2)

!

(lnย 2)

|

|ย |

|ย |

|

(

1)!

(lnย 2)

1

n

n

n

n

n

n

a

x

n

x

a

n

x

n

+

+

+

=

๏ƒ—

=

+

+

ย 

(lnย 2)

limย |

|ย 0

1

n

x

n

โ†’๏‚ฅ

=

+

ย 

Theย limitย is <1ย forย allย xย soย itย converges forย allย realย numbers andย theย radius ofย convergesย is

4

(

0)

)ย /ย (6

1)

0

k

lim

k

k

+

+ย =

ย 

ย 

๐‘…ย =ย โˆž

ย 

ย 

background image

Questionย #4ย 

Express theย definiteย integralย asย anย infiniteย seriesย andย findย itsย valueย toย withinย anย errorย ofย atย mostย 

ย 

10

โˆ’4

ย ย 

ย 

โˆซย cos(๐‘ฅ

2

)ย ๐‘‘๐‘ฅ

1

0

ย 

ย 

Answer:ย 

Term-by-Termย Differentiationย andย Integrationย 

ย 

๐น(๐‘ฅ)ย =ย โˆ‘

๐‘Ž

๐‘›

(๐‘ฅย โˆ’ย ๐‘)

๐‘›

โˆž

๐‘›=0

ย ย 

has radiusย ofย convergenceย ย 

๐‘…ย >ย 0

ย .ย Thenย Fย is differentiableย onย ย 

(๐‘ย โˆ’ย ๐‘…,ย ๐‘ย +ย ๐‘…)

.ย Furtehermore,ย weย canย 

integrateย andย differentiateย termย by term.ย Forย ย x\in(c-R,c+R)ย ย 

ย 

๐น(๐‘ฅ)ย =ย โˆ‘

๐‘›๐‘Ž

๐‘›

(๐‘ฅย โˆ’ย ๐‘)

๐‘›โˆ’1

โˆž

๐‘›=1

ย ย 

ย 

โˆซย ๐น(๐‘ฅ)๐‘‘๐‘ฅย =ย ๐ดย +ย โˆ‘

๐‘Ž

๐‘›

๐‘›+1

(๐‘ฅย โˆ’ย ๐‘)

๐‘›+1

โˆž

๐‘›=0

ย (Aย any constant)ย 

Theseย series haveย theย sameย radius ofย convergenceย Rย 

Hereย weย needย toย findย theย valueย ofย ย 

๐น(๐‘ฅ)ย =ย โˆซย cos(๐‘ฅ

2

)ย ๐‘‘๐‘ฅ

1

0

ย ย weย willย useย theย expansionย ofย theย ย 

cosย ๐‘ฅ

ย ย 

Fromย tableย 2,ย Weย haveย Maclaurinย Series ย 

๐‘“(๐‘ฅ)ย =ย cosย ๐‘ฅย =ย โˆ‘

(โˆ’1)

๐‘›ย ๐‘ฅ

2๐‘›

(2๐‘›)!

โˆž

๐‘›=0

=ย 1ย โˆ’

๐‘ฅ

2

2!

+

๐‘ฅ

4

4!

โˆ’

๐‘ฅ

6

6!

+ย โ‹ฏ

ย ย 

convergesย forย forย allย xย hereย xย isย replacedย withย ย 

๐‘ฅ

2

ย ย 

ย 

cosย ๐‘ฅ

2

=ย โˆ‘

(โˆ’1)

๐‘›ย (๐‘ฅ

2

)

2๐‘›

(2๐‘›)!

โˆž

๐‘›=0

ย ย 

ย 

cosย ๐‘ฅ

2

=ย โˆ‘

(โˆ’1)

๐‘›ย ๐‘ฅ

4๐‘›

(2๐‘›)!

๐‘‘๐‘ฅ

โˆž

๐‘›=0

ย ย 

ย 

โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย =ย โˆ‘

๐‘–๐‘›๐‘“๐‘ก(โˆ’1)

๐‘›

1

(4๐‘›+1)(2๐‘›)!

\
๐‘›=0

ย ย 

Nowย weย needย toย findย outย F(1)ย errorย less thanย 0.0001ย 

ย 

โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย =ย โˆ‘

(โˆ’1)

๐‘›

1

(4๐‘›+1)(2๐‘›)!

โˆž

๐‘›=0

ย ย 

Aboveย isย alternatingย seriesย withย ย 

๐‘Ž

๐‘›

=

1

(4๐‘›+1)(2๐‘›)!

ย ย 

Usingย theย errorย theย boundedย ofย alternatingย seriesย ย 

|โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย โˆ’ย ๐‘†

๐‘

|ย <ย ๐‘Ž

๐‘+1

ย 

ย 

๐‘Ž

๐‘+1

=

1

(4๐‘+5)(2๐‘+2)!

<ย 0.0001

ย ย 

ย 

โ‡’ย (4๐‘ย +ย 5)(2๐‘ย +ย 2)!ย >ย 10000

ย ย 

forย N=2ย weย haveย ย 

(4๐‘ย +ย 5)(2๐‘ย +ย 2)!ย =ย 13ย ร—ย 6!ย =ย 9360ย <ย 10000

ย ย 

forย N=3ย weย haveย ย 

(4๐‘ย +ย 5)(2๐‘ย +ย 2)!ย =ย 17ย ร—ย 8!ย =ย 685440ย >ย 10000

ย ย soย weย chooseย N=3ย 

ย 

โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย โ‰ˆย โˆ‘

๐‘›

=ย 0

โˆž

(โˆ’1)

๐‘›

1

(4๐‘›+1)(2๐‘›)!

ย ย 

โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย =ย 1ย โˆ’

1

5ย ร—ย 2!

+

1

9ย ร—ย 4!

โˆ’

1

13ย ร—ย 6!

ย 

ย 

โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย =ย 0.904522792

ย ย 

Result:ย ย 

โˆซย cosย ๐‘ฅ

2

1

0

๐‘‘๐‘ฅย =ย 0.904522792

ย 

ย 

Questionย #5ย 

Approximateย theย sumย ofย theย series correctย toย fourย decimalย places.ย 

background image

ย 

โˆ‘ย ๐‘›

โˆž

๐‘›

=

1(โˆ’1)

๐‘›

3

๐‘›

๐‘›!

ย ย 

Answer:ย 

Weย knowย that,ย inย anย alternatingย series,ย theย estimateย errorย is smallerย thanย theย firstย neglectedย term.ย That'sย 

whyย weย willย apply.ย Weย canย seeย thatย weย have:ย 

ย 

|๐‘…

๐‘›

|ย โ‰คย ๐‘

๐‘›+1

=

1

3

๐‘›+1

(๐‘›+1)!

<ย 0.00001

ย ย 

Sinceย this canย beย aย toughย inequalityย toย solve,ย weย canย useย graphingย calculatorย toย figureย outย forย whichย nย thisย 

willย aply.ย Weย canย seeย thatย weย have:ย 

ย 

1

3

5

5!

โ‰ˆย 0.00003ย >ย 0.00001ย >

1

3

6

6!

โ‰ˆย 0.000002

ย ย 

Therefore,ย weย needย toย addย atย leastย 5ย terms forย theย errorย toย beย less thanย ย 

0.00001

ย ,ย thatย is,ย forย ourย 

approximationย toย beย correctย toย theย fourthย decimal.ย Nowย weย have:ย 

ย 

๐‘ 

5

=ย โˆ‘

(โˆ’1)

๐‘˜

3

๐‘˜

๐‘˜!

5

๐‘˜=1

โ‰ˆย โˆ’0.28347

ย ย 

Theย sumย ofย theย seriesย isย approximatelyย 

โˆ’0.28347

ย 

Questionย #6ย 

findย aย formulaย forย anย forย theย arithmetic sequence.ย ย 

๐‘Ž

1

=ย โˆ’4,ย ๐‘Ž

5

=ย 16

ย 

ย 

Answer:ย 

Generic formulaย forย anย arithmetic sequenceย inย whichย nthย termย equals theย firstย termย plus theย commonย 

differenceย times theย numberย ofย differences betweenย theย firstย termย andย theย nthย term:ย 

ย 

๐‘Ž

๐‘›

=ย ๐‘Ž

1

+ย ๐‘‘(๐‘›ย โˆ’ย 1)

ย ,ย thenย 

1)ย ย 

๐‘Ž

5

=ย ๐‘Ž

1

+ย 4

๐‘‘

ย ย 

ย 

16ย =ย โˆ’4ย +ย 4

๐‘‘

ย ย 

ย 

๐‘‘ย ย =ย ย 5

ย ย 

2)ย ย 

๐‘Ž

๐‘›

=ย โˆ’4ย +ย 5(๐‘›โ€“ย 1)

ย ย 

ย 

๐‘Ž

๐‘›

=ย โˆ’4ย +ย 5

๐‘›

โ€“ย 5ย =ย 5

๐‘›

โˆ’ย 9

ย ย 

ย 

Questionย #7ย 

Writeย theย firstย fiveย terms ofย theย arithmetic sequence.ย ย 

๐‘Ž

1

=ย 5,ย ๐‘‘ย =ย 6

ย 

ย 

background image

Answer:ย 

Toย findย 5ย terms,ย addย 6ย toย eachย term.ย 

ย 

๐‘Ž

2

=ย ๐‘Ž

1

+ย ๐‘‘ย =ย 5ย +ย 6ย =ย 11,ย ๐‘Ž

1

=ย 5;ย ๐‘‘ย =ย 6

ย ย 

ย 

๐‘Ž

3

=ย ๐‘Ž

2

+ย ๐‘‘ย =ย 11ย +ย 6ย =ย 17,ย ๐‘Ž

2

=ย 11;ย ๐‘‘ย =ย 6

ย ย 

ย 

๐‘Ž

4

=ย ๐‘Ž

3

+ย ๐‘‘ย =ย 17ย +ย 6ย =ย 23,ย ๐‘Ž

3

=ย 17;ย ๐‘‘ย =ย 6

ย ย 

ย 

๐‘Ž

5

=ย ๐‘Ž

4

+ย ๐‘‘ย =ย 23ย +ย 6ย =ย 29,ย ๐‘Ž

4

=ย 23;ย ๐‘‘ย =ย 6

.ย ย 

Theย answerย is:ย 5,ย 11,ย 17,ย 23,ย 29.

ย 

Questionย #8ย 

Whichย ofย theย followingย seriesย areย geometric?ย Whichย areย powerย ones?ย 

a)ย ย 

1ย +ย ๐‘ฅ/2ย +ย ๐‘ฅ

2

/4ย +ย ๐‘ฅ

3

/8ย +ย ๐‘ฅ

4

/16ย +ย โ‹ฏ

ย ย 

b)ย ย 

1ย ย +ย ย 1.1ย ย +ย ย 1.21ย ย +ย ย 1.331ย ย +ย ย 1.4641ย ย +ย ย 1.6105ย ย +ย ย ย โ€ฆย 

ย 

c)ย ย 

(1/3)

2

+ย (1/3)

4

+ย (1/3)

6

+ย (1/3)

8

+ย โ‹ฏ

ย ย 

d)ย ย 

1ย +ย ๐‘ฅย +ย ๐‘ฅ

2

/(2!)ย +ย ๐‘ฅ

3

/(3!)ย +ย ๐‘ฅ

4

/(4!)ย +ย โ‹ฏ

ย ย 

e)ย ย 

1ย ย +ย ย 1/2ย ย +ย ย 1/3ย ย +ย ย 1/4ย ย +ย ย 1/5ย ย +ย ย โ€ฆ

ย ย 

f)ย ย 

1/๐‘ฅ

2

+ย 1/๐‘ฅย +ย 1ย +ย ๐‘ฅย +ย ๐‘ฅ

2

+ย ๐‘ฅ

3

+ย ๐‘ฅ

4

+ย โ‹ฏ

ย 

Answer:ย 

a)ย ย 

1ย +ย ๐‘ฅ/2ย +ย ๐‘ฅ

2

/4ย +ย ๐‘ฅ

3

/8ย +ย ๐‘ฅ

4

/16ย +ย โ‹ฏ

ย ย is bothย powerย andย geometric series.ย 

b)ย 

1ย ย +ย ย 1.1ย ย +ย ย 1.21ย ย +ย ย 1.331ย ย +ย ย 1.4641ย ย +ย ย 1.6105ย ย +ย ย ย โ€ฆ

ย ย is only powerย seriesย 

c)ย 

(1/3)

2

+ย (1/3)

4

+ย (1/3)

6

+ย (1/3)

8

+ย โ‹ฏ

ย ย is geometric series.ย 

d)ย ย 

1ย +ย ๐‘ฅย +ย ๐‘ฅ

2

/(2!)ย +ย ๐‘ฅ

3

/(3!)ย +ย ๐‘ฅ

4

/(4!)ย +ย โ‹ฏ

ย ย is geometric series.ย 

e)ย ย 

1ย ย +ย ย 1/2ย ย +ย ย 1/3ย ย +ย ย 1/4ย ย +ย ย 1/5ย ย +ย ย โ€ฆ

ย ย is neitherย geometric norย powerย one.ย 

f)ย ย 

1/๐‘ฅ

2

+ย 1/๐‘ฅย +ย 1ย +ย ๐‘ฅย +ย ๐‘ฅ

2

+ย ๐‘ฅ

3

+ย ๐‘ฅ

4

+ย โ‹ฏ

ย ย is powerย series.

ย 

Questionย #9ย 

Theย followingย series ย 

โˆ‘

๐‘œ1/๐‘˜

5

๐‘œ

(๐‘˜=0)

ย ย is:ย 

a)ย alternatingย seriesย 

b)ย convergentย p-seriesย 

c)ย divergentย p-seriesย 

d)ย geometric series

ย 

Answer:ย 

ย 

โˆ‘

๐‘œ1/๐‘˜

5

๐‘œ

(๐‘˜=0)

ย ย 

This seriesย is convergentย p-series.ย 

Sinceย ย 

โˆ‘

๐‘œ1/๐‘˜

5

๐‘œ

(๐‘˜=0)

ย converges forย ย 

๐‘ย ย >ย ย 1

ย ย andย divergesย forย ย 

๐‘ย ย โ‰คย ย 1ย 

ย 

Soย compareย this then:ย ย 

๐‘ย ย =ย ย 5ย ย >ย ย 1

ย ย 

Thus,ย givenย series converges by p-series test.

ย 

Questionย #10ย 

background image

Considerย twoย series ย 

โˆ‘

๐‘œ1/2๐‘๐‘–๐‘›ย 

๐‘œ

(๐‘›=1)

and

ย โˆ‘

๐‘œ1/๐‘๐‘–

2

๐‘œ

(๐‘›=0)

๐‘›

.ย Whichย ofย themย converges?

ย 

Answer:ย 

Series A:ย ย 

โˆ‘

๐‘œ1/2๐‘๐‘–๐‘›

๐‘œ

(๐‘›=1)

=ย 1/2๐‘๐‘–1/1ย +ย 1/2ย +ย 1/3ย +

ย ย 

Harmonic seriesย 

โ€“ย Divergentย 

Series B:

โˆ‘

๐‘œ1/๐‘๐‘–

2

๐‘œ

(๐‘›=0)

๐‘›ย =ย (1/๐‘๐‘–

2

)ย +ย (1/๐‘๐‘–

2

)

2

+

ย 

Geometric series,ย soย ย 

1/๐‘๐‘–

2

<ย 1

ย ย 

โ€“ย Convergentย 

Thus,ย only series Bย converges.

ย 

Questionย #11ย 

Nameย theย series:ย 

a)ย ย 

โˆ‘(๐‘›ย =ย 0)

๐‘œ

๐‘œ(๐‘“

๐‘›

๐‘ฅ๐‘ฅ๐‘Ž)/(๐‘›!)๐‘ฅ๐‘ฅ(๐‘ฅโ€“ย ๐‘Ž)

๐‘›

ย ย 

b)ย ย 

โˆ‘(๐‘›ย =ย 0)

๐‘œ

๐‘œ(๐‘“

๐‘›

๐‘ฅ๐‘ฅ0)/(๐‘›!)๐‘ฅ๐‘ฅ๐‘ฅ

๐‘›

ย ย 

Answer:ย 

a)ย ย 

โˆ‘(๐‘›ย =ย 0)

๐‘œ

๐‘œ(๐‘“

๐‘›

๐‘ฅ๐‘ฅ๐‘Ž)/(๐‘›!)๐‘ฅ๐‘ฅ(๐‘ฅโ€“ย ๐‘Ž)

๐‘›

ย ย is aย Taylorย series.ย 

b)ย ย 

โˆ‘(๐‘›ย =ย 0)

๐‘œ

๐‘œ(๐‘“

๐‘›

๐‘ฅ๐‘ฅ0)/(๐‘›!)๐‘ฅ๐‘ฅ๐‘ฅ

๐‘›

ย ย is aย Maclaurinย series.ย 

Questionย #12ย 

Classifyย theย followingย series:ย 

ย 

1ย ย +ย ย 1/4ย ย +ย ย 1/9ย ย +ย ย 1/16ย ย +ย ย 1/25ย ย +ย ย ย โ€ฆ

ย 

ย 

Answer:ย 

ย 

1ย ย +ย ย 1/4ย ย +ย ย 1/9ย ย +ย ย 1/16ย ย +ย ย 1/25ย ย +ย ย ย โ€ฆ

ย ย 

ย 

1/1

2

+ย 1/2

2

+ย 1/3

2

+ย 1/4

2

+ย 1/5

2

+ย โ‹ฏ

ย ,ย soย ย 

โˆ‘

๐‘œ1/(๐‘›)

2

๐‘œ

(๐‘›=1)

ย ย 

Harmonic series areย givenย as:ย ย 

โˆ‘

๐‘œ1/(๐‘›)

๐‘

๐‘œ

(๐‘›=1)

ย ,ย inย theย aboveย seriesย pย = 2ย 

Hence,ย theย series wouldย beย classifiedย as aย p-seriesย withย 

๐‘ย ย >ย ย 1

ย 

Questionย #13ย 

Classifyย whetherย this series convergesย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ((โˆ’1)

๐‘˜

๐‘ฅ๐‘ฅ๐‘˜)/(6๐‘˜

4

+ย 1)

ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ((โˆ’1)

๐‘˜

๐‘ฅ๐‘ฅ๐‘˜)/(6๐‘˜

4

+ย 1)

ย ย 

ย 

4

(ย /ย (6

1)

k

a

k

k

=

+

ย 

ย 

4

(ย /ย (6

1)

k

b

k

k

=

+

ย 

1)ย ย 

๐‘

๐‘˜

โ‰ฅย 0

ย ย 

2)ย ย 

4

(

0)

)ย /ย (6

1)

0

k

lim

k

k

+

+ย =

ย 

Then,ย byย alternativeย seriesย testย weย canย converge.ย 

background image

Theย termย ofย series areย alternativeย andย limitย ofย absoluteย valueย is 0,ย soย series convergeย by alternativeย seriesย 

test.

ย 

Questionย #14ย 

Findย theย sumย ofย theย series,ย ifย itย converges:ย 

a)ย ย 

โˆ‘(๐‘˜ย =ย 0)

๐‘œ

๐‘œ(1/3)

๐‘˜

ย ย 

b)ย ย 

โˆ‘(๐‘˜ย =ย 0)

๐‘œ

๐‘œ(โˆ’1/2)

๐‘˜

ย 

ย 

Answer:ย 

a)ย ย 

โˆ‘(๐‘˜ย =ย 0)

๐‘œ

๐‘œ(1/3)

๐‘˜

ย ย 

Aย seriesย ofย formย ย 

๐‘Ÿ

๐‘˜

ย ย converges forย ย 

|๐‘Ÿ|ย <ย 1

ย ย 

Here,ย ย 

๐‘Ÿย ย =ย ย 1/3ย ย <ย ย 1

ย ,ย soย theย series converges.ย 

ย 

โˆ‘(๐‘˜ย =ย 0)

๐‘œ

๐‘œ(1/3)

๐‘˜

=ย 1/(1โ€“ย (1/3))ย =ย 1/(2/3)ย =ย 3/2

ย ย 

b)ย ย 

โˆ‘(๐‘˜ย =ย 0)

๐‘œ

๐‘œ(โˆ’1/2)

๐‘˜

ย ย 

Here,ย ย 

๐‘Ÿย ย =ย ย ย โˆ’1/2

ย ,ย soย theย series converges.ย 

ย 

โˆ‘(๐‘˜ย =ย 0)

๐‘œ

๐‘œ(โˆ’1/2)

๐‘˜

=ย 1/(1โ€“ย (โˆ’1/2))ย =ย 1/(1ย +ย 1/2)ย =ย 2/3

ย ย 

ย 

ย 

Questionย #15ย 

Determineย whetherย theย series converges:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ4/(๐‘˜ย +ย 2)

5

ย 

ย 

background image

Answer:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ4/(๐‘˜ย +ย 2)

5

ย ย 

ย 

๐‘Ž

๐‘˜

=ย 4/(๐‘˜ย +ย 2)

5

ย ย 

ย a_k = 4/(k^5(1ย + 2/k)^5)ย ย 

Letย assumeย that,ย 

ย 

๐‘

๐‘˜

=ย 1/๐‘˜

5

ย ย 

By theย limitย comparisonย test,ย 

ย 

๐ฟย =ย ๐‘™๐‘–๐‘š

(๐‘˜โ†’๐‘œ๐‘œ)

๐‘Ž

๐‘˜

/๐‘

๐‘˜

ย ย 

ย 

๐ฟย =ย ๐‘™๐‘–๐‘š

(๐‘˜โ†’๐‘œ๐‘œ)

(4/(๐‘˜

5

(1ย +ย 2/๐‘˜)

5

))/(1/๐‘˜

5

)

ย ย 

ย 

๐ฟย =ย ๐‘™๐‘–๐‘š

(๐‘˜โ†’๐‘œ๐‘œ)

4/(1ย +ย 2/๐‘˜)

5

ย ย 

ย 

๐ฟย =ย 4(1ย +ย 2/๐‘œ๐‘œ)

5

ย ย 

ย 

๐ฟย =ย 4/(1ย +ย 0)

5

ย ย 

ย 

๐ฟย ย =ย ย 4

ย ย 

Hereย ย 

๐ฟย ย =ย ย 4

ย ย 

โ€“ย finiteย andย non-zero,ย thereforeย theย seriesย is aย p-seriesย withย ย 

๐‘ย =ย 5(๐‘

๐‘˜

=ย 1/๐‘˜

5

)

ย ,ย soย theย givenย 

seriesย is converges byย theย properties ofย p-series.ย 

ย 

ย 

background image

Questionย #16ย 

Determineย whetherย theย followingย seriesย converges:ย ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ(2/(๐‘˜ย +ย 5)

3

)

ย .ย 

Answer:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ(2/(๐‘˜ย +ย 5)

3

)

ย ย 

ย 

lim

(๐‘˜โ†’๐‘œ๐‘œ)

(๐‘Ž

๐‘˜

)ย =

lim

((๐‘˜โ†’5)

5

)

(2/(๐‘˜ย +ย 5)

3

)ย =ย 9/๐‘œ๐‘œย =ย 0

ย ย 

Theย limitย ofย theย terms ofย theย seriesย is 0,ย soย theย seriesย diverges by theย divergentย test.

ย 

Questionย #17ย 

Determineย whetherย theย followingย seriesย convergesย ifย โ€œaโ€ย is real:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ(1โ€“ย ๐‘Ž/๐‘˜)

6

๐‘˜

ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ(1โ€“ย ๐‘Ž/๐‘˜)

6

๐‘˜

ย ย 

ifย ย 

lim

(๐‘›โ†’๐‘œ๐‘œ)

๐‘Ž

๐‘›

!ย =ย 0

,ย thenย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ๐‘Ž

๐‘˜

ย ย diverges.ย 

Hereย ย 

๐‘Ž

๐‘˜

=ย (1โ€“ย ๐‘Ž/๐‘˜)

(๐‘˜/6)

ย 

Soย ย 

lim

(๐‘˜โ†’๐‘œ๐‘œ)

๐‘Ž

๐‘˜

=

ย ย 

ย 

=ย lim

(๐‘˜โ†’๐‘œ๐‘œ)

(1โ€“ย ๐‘Ž/๐‘˜)

(๐‘˜/6)

=

ย ย 

ย 

=

lim

(๐‘’

(๐‘˜โ†’๐‘œ๐‘œ)

)

(๐‘˜/6)ย ๐‘ฅ๐‘ฅ(โˆ’๐‘Ž/๐‘˜)ย =

ย ย 

ย 

=

lim

(๐‘’

(๐‘˜โ†’๐‘œ๐‘œ)

)

โˆ’ย ๐‘Ž/6ย =

ย ย 

ย 

lim

(๐‘˜โ†’๐‘œ๐‘œ)

๐‘Ž

๐‘˜

=ย ๐‘’

(โˆ’๐‘Ž/6)

ย ย 

aย isย realย number,ย soย ย 

๐‘’

(โˆ’๐‘Ž/6)

!ย =ย 0

ย ย forย realย numberย a.ย 

ย 

lim

(๐‘˜โ†’๐‘œ๐‘œ)

๐‘Ž

๐‘˜

!ย =ย 0

ย ย 

Thus,ย byย divergentย test,ย givenย seriesย is divergent.

ย 

Questionย #18ย 

Determineย whetherย theย followingย seriesย converges orย diverges:ย 

ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(๐‘›

3

+ย ๐‘›

4

)/๐‘›

5

.ย 

ย 

Answer:ย 

ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(๐‘›

3

+ย ๐‘›

4

)/๐‘›

5

=

ย ย 

ย 

=ย โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ1/๐‘›

2

+ย โˆ‘

๐‘œ1/๐‘›

๐‘œ

(๐‘›=1)

ย 

As ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ1/๐‘›

2

ย ย is convergentย by p-series testย andย 

โˆ‘

๐‘œ1/๐‘›

๐‘œ

(๐‘›=1)

ย ย isย divergent,ย 

Seriesย divergesย becauseย weย canย break theย seriesย upย intoย aย convergentย p-series andย aย divergentย p-series.

ย 

Questionย #19ย 

Determineย whetherย theย followingย seriesย converges orย diverges:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ3/(๐‘˜ย +ย 4)

3

ย 

ย 

background image

Answer:ย 

ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ3/(๐‘˜ย +ย 4)

3

=

ย ย 

ย 

=ย โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ3/๐‘˜

3

ย ย 

Byย p-series testย ย 

โˆ‘(๐‘˜ย =ย 1)

๐‘œ

๐‘œ1/โ„Ž

๐‘

,ย ๐‘ย >ย 1

ย 

โ€“ย converges,ย ย 

๐‘ย ย <ย ย 1

ย ย 

โ€“ย diverges.ย 

Soย this series converges,ย causeย itโ€™s aย p-seriesย withย pย = 3.

ย 

Questionย #20ย 

Please,ย describeย theย applicationย ofย theย Alternatingย Series Testย forย eachย series:ย 

a)ย ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(โˆ’1)

๐‘›

/๐‘›

ย ย 

b)ย ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ1/๐‘›

ย ย 

c)ย ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(โˆ’1)

๐‘›

ย 

ย 

Answer:ย 

a)ย ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(โˆ’1)

๐‘›

/๐‘›

ย ย 

โ€“ย theย givenย series convergesย 

b)ย ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ1/๐‘›

ย ย 

โ€“ย theย Alternatinย Seriesย Testย isย inconclusiveย orย cannotย beย appliedย toย theย seriesย 

c)ย ย 

โˆ‘(๐‘›ย =ย 1)

๐‘œ

๐‘œ(โˆ’1)

๐‘›

ย ย 

โ€“ย theย Alternatinย Series Testย isย inconclusiveย orย cannotย beย appliedย toย theย seriesย 

ย 

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